How do we maximise the effectiveness of our rerolls? I’ve got a feeling that this one might end up stating the obvious a little, but I like the maths, so whatever.
As a side project, I’m running my brood brothers contingent as legitimate Guard – specifically, Guard with the Gunnery Experts doctrine, which allows me to reroll the number of shots for any variable shot weapon (like double-tapping battle-cannons, for example). So how can we maximise the effectiveness of a reroll, and what does that do to the expected outcomes, especially across two dice?
As it turns out, we can push the average of a single die roll up from 3.5 to 4.25 by holding onto 4, 5 and 6, and rerolling 1, 2 and 3. Bluntly, reroll anything that comes in under the original average, and stick with anything over it. Which makes sense.
[Incidentally, there’s a neat symmetry around that ideal outcome, with the averages dropping away in pairs on either side – you get the same average by rerolling 2s or less as you do rerolling 4s or less (4.17); and then the same again for rerolling 1s and rerolling 5s or less (3.92). So that’s nice.]
And in the cases where we can apply this to both parts of a two-dice roll, the averages simply double. A battle cannon with Gunnery Experts will average 8.5 (rather than 7) if it can fire twice and rerolls 3s or less.
EXPECTED OUTCOMES
The more interesting part of this is what it does to expected outcomes.
The most likely outcome of a straight two-dice roll is 7. It comes up 1 in 6 rolls, or 17% of the time. But once we start rerolling, the most likely outcome changes.
· Rerolling just 1s makes the most likely outcome an 8 (19% chance). It also gives a better than 50% chance of scoring 8 or better overall.
· Rerolling 2 or less makes the most likely outcome 9 (20% chance). It gives a 67% (two thirds) chance or scoring 8 or more, and almost a 50% chance of 9 or more (falling just shy at 49%).
· Rerolling 3 or less [which generates the highest average for a single die] makes the most likely outcome 10 (19% chance), and gives a 54% chance or scoring 9 or better.
· Rerolling 4 or less is odd, because 11 and 7 are equally likely (both 15%). It gives a 63% chance of 8 or more, and a 49% chance of 9 or more (which is similar to rerolling 2s or less).
· Rerolling 5 or less makes 7 the most likely outcome (16% chance – slightly less than not rerolling at all), and gives a 55% chance of scoring 8 or more overall.
There’s a couple of other interesting bits and pieces.
Each set of rerolls has a better chance of producing a certain number than any of the others. The best odds for rolling a 12 comes from rerolling 5s or less; an 11, 4s or less; 10, 3s or less; 9, 2s or less; 8, reroll 1s; and the best odds on rolling a 7 (or anything less than 7) come from not rerolling at all. This all makes sense, given that it’s a long-winded way of saying ‘reroll anything that makes your target impossible’ – if you’re after a 12, you need to reroll anything that’s not a 6 (so, 5s or less); if you want a 10, you need both dice to be at least 4, so reroll any 3s or less.
Which is pretty maths, but not all that useful – we rarely ever want a specific number across two dice (or across a single roll, for that matter). What interests us more is the chance of getting any particular value or better.
THE POTENTIALLY USEFUL BIT
This pattern starts the same way: if you want a 12, reroll anything that isn’t a 6; for 11+, reroll 4s or less; and for 10+, reroll 3s or less. But then you want to stick with that for 9, 8 and 7. It turns out that rerolling 3 or less gives you the best odds of anything from 7 to 10, with an 81% chance of making at least the 7, and a 38% chance of 10+. (By way of comparison, rolling a 5+ on a single die is 34%, and getting a 2+ is 83%, which feels like good odds – if someone offered me the chance to make 7” charges on a 2+, I’d take it. To be honest, I’d take 10” charges on a 5+ too, given that it’s double the usual odds.)
(Below 7, you’re slightly better of holding onto your 3s (so that you don’t turn them into 1s or 2s), but it’s marginal. And by this point, it’s less important, as it tends to be the top end we’re interested in most of the time.)
The thing is, the practicalities of this break down in real life, but the maths still applies. Saying that you get the best results by holding 4+ and rerolls 3s or less is fine until you roll a pair of 4s when you need 9. Then what? If we run the odds, we see that holding one 4 and rerolling the other gives a 33% chance of getting the 5 or 6 you need; rerolling both gives a 28% of getting a 9+. Better to hold onto one of the 4s.
If we'd wanted 10+, we're also no worse off holding one of the 4s - a single die needing a 6 has the same chance of coming good as two dice looking for a natural 10+.
But what about lower numbers? What if we needed a 7, but rolled double 3s? Rerolling just one die gives a 50% chance of producing the 4, 5 or 6 we need. But rerolling both gives a 58% chance of hitting a 7+. So sticking to 'reroll 3s or less' remains a useful maxim.
Similarly, if we want 6 but roll 2 and 3, should we hold the 3? No. Rerolling just the 2 has a 67% chance of getting the 3+ you'd need to reach 6. But rerolling both has a 72% or getting a 6+ anyway. On the other hand, if we'd rolled 4 and 1, we should hold the 4 and reroll the 1, because we'd only need a 2+, and a single die gets that 83% of the time (which is this case is better odds than rerolling both).
CONCLUSION
I know. This is a niche bit of maths. There aren’t many situations where we can selectively reroll a 2d6 roll on both/either/neither basis – the Gunnery Experts example I started off with, possibly Orks on the charge, maybe a couple of other things. So this isn’t massively practical. But it does answer my question – the best way to maximise a d6 roll (if we’re just looking for more, rather than something specific) is to hold 4, 5 and 6, and reroll the rest. As I said at the top, this feels like the obvious conclusion. But I guess it's good to know that I'm not just running off a hunch, and the numbers really do back it up.